计算两个解决方案之间差值的无穷大范数

Question

在下面的代码中，我已经能够：

1. 对一般平方线性系统实现高斯消除，无需枢轴。
2. 我已经通过解决 Ax=b 对其进行了测试，其中 A 是随机的 100x100 矩阵，b 是随机的 100x1 向量。
3. 我已经将我的解决方案与使用 numpy.linalg.solve 获得的解决方案进行了比较

但是，在最后一项任务中，我需要计算两个解决方案之间差值的无穷大范数。我知道无穷大范数是矩阵的最大绝对行和。但是我怎么能做到这一点来计算两个解决方案之间差异的无穷大范数，我的解决方案和 numpy.linalg.solve。寻求一些帮助！

import numpy as np
def GENP(A, b):
'''
Gaussian elimination with no pivoting.
% input: A is an n x n nonsingular matrix
%        b is an n x 1 vector
% output: x is the solution of Ax=b.
% post-condition: A and b have been modified. 
'''
n =  len(A)
if b.size != n:
raise ValueError("Invalid argument: incompatible sizes between A & b.", b.size, n)
for pivot_row in range(n-1):
for row in range(pivot_row+1, n):
multiplier = A[row][pivot_row]/A[pivot_row][pivot_row]
#the only one in this column since the rest are zero
A[row][pivot_row] = multiplier
for col in range(pivot_row + 1, n):
A[row][col] = A[row][col] - multiplier*A[pivot_row][col]
#Equation solution column
b[row] = b[row] - multiplier*b[pivot_row]
x = np.zeros(n)
k = n-1
x[k] = b[k]/A[k,k]
while k >= 0:
x[k] = (b[k] - np.dot(A[k,k+1:],x[k+1:]))/A[k,k]
k = k-1
return x
if __name__ == "__main__":
A = np.round(np.random.rand(100, 100)*10)
b =  np.round(np.random.rand(100)*10)
print (GENP(np.copy(A), np.copy(b)))

例如，此代码为上面列出的任务 1 提供以下输出：

[-6.61537666  0.95704368  1.30101768 -3.69577873 -2.51427519 -4.56927017
-1.61201589  2.88242622  1.67836096  2.18145556  2.60831672  0.08055869
-2.39347903  2.19672137 -0.91609732 -1.17994959 -3.87309152 -2.53330865
5.97476318  3.74687301  5.38585146 -2.71597978  2.0034079  -0.35045844
0.43988439 -2.2623829  -1.82137544  3.20545721 -4.98871738 -6.94378666
-6.5076601   3.28448129  3.42318453 -1.63900434  4.70352047 -4.12289961
-0.79514656  3.09744616  2.96397264  2.60408589  2.38707091  8.72909353
-1.33584905  1.30879264 -0.28008339  0.93560728 -1.40591226  1.31004142
-1.43422946  0.41875924  3.28412668  3.82169545  1.96675247  2.76094378
-0.90069455  1.3641636  -0.60520103  3.4814196  -1.43076816  5.01222382
0.19160657  2.23163261  2.42183726 -0.52941262 -7.35597457 -3.41685057
-0.24359225 -5.33856181 -1.41741354 -0.35654736 -1.71158503 -2.24469314
-3.26453092  1.0932765   1.58333208  0.15567584  0.02793548  1.59561909
0.31732915 -1.00695954  3.41663177 -4.06869021  3.74388762 -0.82868155
1.49789582 -1.63559124  0.2741194  -1.11709237  1.97177449  0.66410154
0.48397714 -1.96241854  0.34975886  1.3317751   2.25763568 -6.80055066
-0.65903682 -1.07105965 -0.40211347 -0.30507635]

然后对于任务二，我的代码给出以下内容：

my_solution = GENP(np.copy(A), np.copy(b))
numpy_solution = np.linalg.solve(A, b)
print(numpy_solution)

导致：

[-6.61537666  0.95704368  1.30101768 -3.69577873 -2.51427519 -4.56927017
-1.61201589  2.88242622  1.67836096  2.18145556  2.60831672  0.08055869
-2.39347903  2.19672137 -0.91609732 -1.17994959 -3.87309152 -2.53330865
5.97476318  3.74687301  5.38585146 -2.71597978  2.0034079  -0.35045844
0.43988439 -2.2623829  -1.82137544  3.20545721 -4.98871738 -6.94378666
-6.5076601   3.28448129  3.42318453 -1.63900434  4.70352047 -4.12289961
-0.79514656  3.09744616  2.96397264  2.60408589  2.38707091  8.72909353
-1.33584905  1.30879264 -0.28008339  0.93560728 -1.40591226  1.31004142
-1.43422946  0.41875924  3.28412668  3.82169545  1.96675247  2.76094378
-0.90069455  1.3641636  -0.60520103  3.4814196  -1.43076816  5.01222382
0.19160657  2.23163261  2.42183726 -0.52941262 -7.35597457 -3.41685057
-0.24359225 -5.33856181 -1.41741354 -0.35654736 -1.71158503 -2.24469314
-3.26453092  1.0932765   1.58333208  0.15567584  0.02793548  1.59561909
0.31732915 -1.00695954  3.41663177 -4.06869021  3.74388762 -0.82868155
1.49789582 -1.63559124  0.2741194  -1.11709237  1.97177449  0.66410154
0.48397714 -1.96241854  0.34975886  1.3317751   2.25763568 -6.80055066
-0.65903682 -1.07105965 -0.40211347 -0.30507635]

最后是任务 3：

if np.allclose(my_solution, numpy_solution):
print("These solutions agree")
else:
print("These solutions do not agree")

导致：

These solutions agree

Answer 1

如果你想要的只是矩阵的无穷范数， 它通常应该看起来像这样：

def inf_norm(matrix):
return max(abs(row.sum()) for row in matrix)

但由于您的my_solution和numpy_solution只是一维向量，因此您 可以重塑它们(我假设 100x1，这就是你所拥有的 示例(与上述功能一起使用：

备选方案1：

def inf_norm(matrix):
return max(abs(row.sum()) for row in matrix)
diff = my_solution - numpy_solution
inf_norm_result = inf_norm(diff.reshape((100, 1))

备选方案2：

或者，如果您知道它们将始终是一维向量，则可以省略总和 (因为所有行的长度均为 1(并直接计算：

abs(my_solution - numpy_solution).max()

备选方案3：

或者按照numpy.linalg.norm(见下文(文档中编写的那样：

max(sum(abs(my_solution - numpy_solution), axis=1))

备选案文4：

或使用numpy.linalg.norm()(请参阅：https://docs.scipy.org/doc/numpy-1.15.1/reference/generated/numpy.linalg.norm.html(：

np.linalg.norm(my_solution - numpy_solution, np.inf)